Please help, will give medal!
What is the taxidistance between (8, 17) and (21, 32)? If AR = 12 m and RE = 3 m, what is the ratio of the areas of triangle ARH and triangle AES?
There should be a figure.
For the second one? Yeah just a second.
For the first one - Use distance formula : If you have 2 points : \(A(x_1,y_1) \space \& \space B (x_2,y_2)\) Then the distance between these two points is : \(AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 } \)
so apparently the first one should be about 17
However, my options are 44 36 28 20
For the second one - Use this theorem : \(\text{In two similar triangles, the ratio of their areas is the square of the ratio of their sides. }\) First of all , you need to prove that the triangles are similar.
@ScarlettBlack , may be you did a mistake in calculation : \( = \sqrt{(21-8)^2 + (32-17)^2 } \) Which I get : 19.849 units or 20 units.
Ok, yes I made a mistake in the second calculation.
Yes. Now, for the second one, can you prove those 2 triangles similar?
For the second one. I thought the ratio was 4:5
Yes, good, the ratio of AR : AE = 4 : 5
But before you use the theorem I provided, you need to prove that the 2 triangles are similar. Do you have any ideas for this?
Well they share a common angle
http://www.ask-math.com/basic-proportionality-theorem.html You will have to use Basic Proportionality Theorem to prove that the ratio of the corresponding sides is equal to the ratio of AR and AE . Then, you can easily state that since the ratios of the corresponding sides is equal , thus, the triangles will be similar. Have you learnt Basic Proportionality Theorem?
Yes.
So, use this theorem.
I actually have to go now. I'm visiting a friend in the hospital. Thanks for the help. I will definitely do the rest.
Oh. Get well soon to your friend. Have a great day ahead! Let me know if you have any doubts.
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